Let A be a dense of [a,b] and let f:[a,b]-->R be an integrable function such that f(x)>=0 for every element of x in A. Prove that the (integral of a to b) f >= 0.

Please help. I have ideas for this but need help solving

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- Apr 1st 2010, 07:06 AMderek walcottDense subset
Let A be a dense of [a,b] and let f:[a,b]-->R be an integrable function such that f(x)>=0 for every element of x in A. Prove that the (integral of a to b) f >= 0.

Please help. I have ideas for this but need help solving - Apr 1st 2010, 08:26 AMTinyboss
I assume you mean f is Riemann-integrable--it isn't true for Lebesgue integration.

For any partition P, every interval will contain an element of A, and therefore the upper sum is non-negative for that partition. Since f is integrable, the upper sums converge to the integral as the mesh goes to zero, and so the integral is non-negative. - Apr 1st 2010, 08:47 AMDrexel28
Another way to say this depending your def. of Riemann integrable. As

**Tinyboss**said $\displaystyle \sup_{x\in I\subseteq[a,b]}f(x)\geqslant 0$. Thus, $\displaystyle U(P,f)\sum_{j=1}^{n}\sup_{x\in[x_{j-1},x_j]}f(x)\Delta x_j\geqslant 0$ regardless of $\displaystyle P$. Clearly then $\displaystyle 0\geqslant U(P,f)$ for every partition $\displaystyle P$ in $\displaystyle \mathcal{P}[a,b]$. Thus, $\displaystyle \int_a^bf=\inf_{P\in\mathcal{P}[a,b]}U(P,f)\geqslant 0$